= Solution
The <polynomial> is <Eisenstein> at every <prime number> $p$, so a <polynomial root> $\alpha$ generates a totally ramified cubic <field> $F/\mathbb Q_p$. The splitting-field <Galois group> is a transitive <subgroup> of $S_3$, hence $C_3$ or $S_3$. Its <polynomial discriminant> is
$$
\Delta=-p^2(4p+27).
$$
The product of the three <polynomial root> differences is a <square root> of $\Delta$, transformed by the sign of the <polynomial root> permutation. Thus the <group> is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the <splitting field> is
$$
L=F\,\mathbb Q_p(\sqrt{\Delta}).
$$
Indeed adjoining one <polynomial root> leaves a quadratic factor, whose <polynomial discriminant> is $\Delta/f'(\alpha)^2$, so the <square root> of $\Delta$ supplies its remaining <polynomial roots>.
Suppose first that $p>3$. The <unit> $-(4p+27)$ reduces to $-27$, and <Hensel's lemma> makes it a square exactly when $-3$ is a square modulo $p$. <Quadratic reciprocity> gives
$$
\left(\frac{-3}{p}\right)=
\begin{cases}1,&p\equiv1\pmod3,\\-1,&p\equiv2\pmod3.\end{cases}
$$
If $p\equiv1\pmod3$, the <field> generated by $\alpha$ is already the cyclic <splitting field>, totally ramified of degree three. If $p\equiv2\pmod3$, the nonsquare <unit> gives an unramified quadratic <field> $E=\mathbb Q_p(\sqrt{\Delta})$. Since $F$ and $E$ have coprime degrees, $[FE:\mathbb Q_p]=6$. Its <residue degree> is at least two and its <ramification index> at least three; their product is six, so they are exactly two and three. Thus its <inertia group> is the normal $C_3$ inside $S_3$.
In both cases the <ramification index> is three and is not divisible by $p$. The <wild inertia group> is trivial, so all positive lower <groups> are trivial. As a different check, $f'(\alpha)=3\alpha^2+p$ has <valuation> two in $F$ when $p>3$, agreeing with tame <different exponent> $3-1$.
For $p=3$ the <polynomial discriminant> is $-351=-3^3\cdot13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E=\mathbb Q_3(\beta)$ with $\beta^2=-39$. It is ramified of degree two, by the <Eisenstein polynomial> $T^2+39$. Together with the totally ramified cubic <field> this forces $e(L/\mathbb Q_3)$ to be divisible by both two and three. Since $[L:\mathbb Q_3]=6$, the <splitting field> is totally ramified. In its normalized <valuation>,
$$
v_L(3)=6,\quad v_L(\alpha)=2,\quad
v_L(\beta)=3,\quad \pi=\beta/\alpha\text{ has }v_L(\pi)=1.
$$
All differences between distinct <polynomial roots> have the same <valuation>, since $S_3$ permutes the <polynomial root> pairs and preserves the <absolute value on a field>. Also
$$
f'(\alpha)=3(\alpha^2+1)
=(\alpha-\alpha_2)(\alpha-\alpha_3)
$$
has <valuation> six, so each <polynomial root> difference has <valuation> three. A nonidentity <three-cycle> fixes $\beta$, and therefore
$$
v_L(\sigma\pi-\pi)
=v_L\left(\frac{\beta(\alpha-\sigma\alpha)}
{\alpha\,\sigma\alpha}\right)
=3+3-2-2=2.
$$
The <transposition> fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement <valuation> $v_L(2\pi)=1$. The other <transpositions> have the same value by conjugacy. The <uniformizer> criterion now determines every <group>:
$$
\boxed{
\begin{array}{c|c|c|c|c}
p&G&e&f&(G_0,G_1,G_2,\ldots)\\
p>3,\ p\equiv1\pmod3&C_3&3&1&(C_3,1,1,\ldots)\\
p>3,\ p\equiv2\pmod3&S_3&3&2&(C_3,1,1,\ldots)\\
p=3&S_3&6&1&(S_3,C_3,1,1,\ldots)
\end{array}}
$$
For the wild case the <different exponent> is $5+2=7$. Independently, the cubic subfield has <different exponent> $v_F(3(\alpha^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes <ramification of the splitting field of X3 plus pX plus p> for every odd <prime number>.
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